sound much as the law of cosines and sines on this one.
sketch it. first you have two sides and angle between to get opposite side. Now use the law of sines to get the last angle, Notice you have to convert that to a bearing relative ot North.
Two ships A and B leaves the some harbour at the same time in directions 152° and 142° respectively calculate the distance between the two ships when A has travelled 10km and B has travelled 12km. What is the bearing of A from B at this time
5 answers
All angles are measured CW from +y-axis.
Given: HA = 10km[152o], HB = 12km[142o].
BH = 12km[142+180] = 12km[322o].
BA = BH + HA = 12[322] + 10[152]
BA = (12*sin322+10*sin152) + (12*cos322+10*cos152)i
BA = -2.69 + 0.627i = 2.76km[-76.9o] = 2.76km[283o] CW.
Bearing of A from B = 283 degrees.
Given: HA = 10km[152o], HB = 12km[142o].
BH = 12km[142+180] = 12km[322o].
BA = BH + HA = 12[322] + 10[152]
BA = (12*sin322+10*sin152) + (12*cos322+10*cos152)i
BA = -2.69 + 0.627i = 2.76km[-76.9o] = 2.76km[283o] CW.
Bearing of A from B = 283 degrees.
The sketch of the journey was not drawn
Why formulary was used here
I'm not sure it's cosine
I'm not sure it's cosine
The answer is 2.765
2 answers